Question

transform y_1=-4sin(t/2)+3cos(t/2) into
y_2=Asin(Bt+C) using y=Msin(Bt)+Ncos(Bt)=
sqrt{N^2+M^2}sin(Bt+C)
we want to choose C such that |C| is a minimum. so i have so far y_2=5sin(t/2+C). Does C need to be zero?

Answer 1

\[y=Msin(Bt)+Ncos(Bt)= \sqrt{N^2+M^2}\sin(Bt+C) \]

Answer 2

compute C to three decimal places
so i guess C shouldn't be zero

Answer 3

and i have figured out that
cos(C)=M/sqrt{M^2+N^2}
and
sin(C)=N/sqrt{M^2+N^2}

Answer 4

so we need cot(C)=-4/3
so this means C =?????
C=-36.870 or -36.87

Answer 5

that was in degrees
i think i will write answer in radians
so i think the answer is y=sqrt{M^2+N^2}sin(t/2-.644)
what do you guys think?
\[y=\sqrt{M^2+N^2}\sin(t/2-.644) \]

Answer 6

oops M=-4 and N=3
so
we have
y=5sin(t/2-.644)